This is for Spivak's Calculus, third edition, chapter 3, problem 7. It says
Prove that for any polynomial function f and any number a,there is polynomial g(x) such that f(x) = (x-a)g(x) + b. Proof is possible by induction on degree of polynomial
Proving for one degree is easy, so I'll skip it. My problem is while proving k+1. The solution that the book provides is
I know it is in Spanish (my native language), but I hope it is basic or obvious enough to understand. Anyhow, my attempt is as follows:
$f(x)=a_0+a_1x+…a_{k+1}x^{k+1} $
and I can get a k degree polynomial (which is less than the original k+1 degree) by subtracting the term $c_{k+1}x^{k+1} $, so we have:
$f(x) – a_{k+1}x^{k+1} = (x−a)g(x)+b $
so far it matches the answers I've found on the internet, but I don't know how to turn that into the answer:
$f(x) – a_{k+1}(x-a) = (x−a)g(x)+b $
So my question is how can I turn
$a_{k+1}x^{k+1} $ into $ a_{k+1}(x-a) $
or am I trying to solve this incorrectly from the beginning?
I am new at proofs and I am studying by myself, so the answer is probably simple, but I have been stuck on this for a while now and I can't figure it out. I appreciate your help.
Best Answer
This looks like a typo. $f(x) - a_{k+1}(x-a)$ should have been $f(x) - a_{k+1}(x-a)^{\color{red}{k+1}}$, otherwise, I don't see how the degree of the polynomial would change for higher power.
Since $f(x) - a_{k+1}(x-a)^{k+1}$ is of degree at most $k$, we can write it as
$$f(x) - a_{k+1}(x-a)^{k+1}=(x-a)g(x)+b$$
$$f(x) =(x-a)[g(x)+a_{k+1}(x-a)^k]+b$$